Question

calculate the energy of a photon of blue light that has a wavelength of 432 nm.
○ 8.59 x 10⁻¹⁷ j
○ 1.53 x 10⁻²⁷ j
○ 4.60 x 10⁻¹⁹ j
○ 8.39 x 10⁻²³ j
○ 1.29 x 10⁶ j

Explanation:

Step1: Recall photon energy formula

The energy \( E \) of a photon is given by \( E = h
u \), and since \( c=\lambda
u \) (where \( c \) is the speed of light, \( \lambda \) is wavelength, \(
u \) is frequency), we can substitute \(
u=\frac{c}{\lambda} \) into the energy formula to get \( E=\frac{hc}{\lambda} \). Here, \( h = 6.626\times 10^{-34}\, \text{J·s} \), \( c = 3.0\times 10^{8}\, \text{m/s} \), and \( \lambda = 432\,\text{nm}=432\times 10^{-9}\, \text{m} \).

Step2: Substitute values into the formula

\[

$$\begin{align*} E&=\frac{(6.626\times 10^{-34}\, \text{J·s})(3.0\times 10^{8}\, \text{m/s})}{432\times 10^{-9}\, \text{m}}\\ &=\frac{1.9878\times 10^{-25}\, \text{J·m}}{432\times 10^{-9}\, \text{m}}\\ &=\frac{1.9878\times 10^{-25}}{4.32\times 10^{-7}}\, \text{J}\\ &\approx 4.60\times 10^{-19}\, \text{J} \end{align*}$$

\]

Answer:

\( 4.60\times 10^{-19}\, \text{J} \) (corresponding to the option with this value)